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Question-217447




Question Number 217447 by Tawa11 last updated on 14/Mar/25
Commented by mr W last updated on 14/Mar/25
question is wrong. the friction   coefficient between cylinder and  stick may not be the same as the  friction coefficient between cylinder  and plane. the former must be  larger than the latter! i think this  can be said even just by using your  feeling without physics calculations.
questioniswrong.thefrictioncoefficientbetweencylinderandstickmaynotbethesameasthefrictioncoefficientbetweencylinderandplane.theformermustbelargerthanthelatter!ithinkthiscanbesaidevenjustbyusingyourfeelingwithoutphysicscalculations.
Commented by Tawa11 last updated on 14/Mar/25
Thanks for your time sir.  I really appreciate.
Thanksforyourtimesir.Ireallyappreciate.
Commented by mr W last updated on 14/Mar/25
where did you have this question?  do they have an answer?
wheredidyouhavethisquestion?dotheyhaveananswer?
Commented by Tawa11 last updated on 14/Mar/25
Sir, I will confirm if there is answer now.
Sir,Iwillconfirmifthereisanswernow.
Commented by Tawa11 last updated on 14/Mar/25
Sir, they got    ((3 sinθ)/(1  +  cosθ))
Sir,theygot3sinθ1+cosθ
Commented by mr W last updated on 14/Mar/25
then the question and diagram are  wrong.  if the question and the diagram were  right, then we had:
thenthequestionanddiagramarewrong.ifthequestionandthediagramwereright,thenwehad:
Commented by mr W last updated on 14/Mar/25
Commented by mr W last updated on 14/Mar/25
the three forces must meet at the  same point as shown. but they are  not in equilibrium!  (with tan φ=μ)
thethreeforcesmustmeetatthesamepointasshown.buttheyarenotinequilibrium!(withtanϕ=μ)
Commented by mr W last updated on 14/Mar/25
in this case, such that equilibrium   is possible, the friction coefficients  must be different. then we have:
inthiscase,suchthatequilibriumispossible,thefrictioncoefficientsmustbedifferent.thenwehave:
Commented by mr W last updated on 14/Mar/25
Commented by mr W last updated on 14/Mar/25
tan φ_1 =μ_1   tan φ_2 =μ_2   θ=2φ_2  ⇒φ_2 =(θ/2)  μ_2 =tan^(−1) φ_2 =tan^(−1) (θ/2)  φ_1 >φ_2  ⇒μ_1 >μ_2   (R_1 /(sin φ_2 ))=(R_2 /(sin φ_1 ))=((Mg)/(sin (φ_1 −φ_2 )))  ⇒R_1 =((sin φ_2 Mg)/(sin (φ_1 −φ_2 )))  ⇒R_2 =((sin φ_1 Mg)/(sin (φ_1 −φ_2 )))  N_1 =R_1 cos φ_1 =((mg)/2)  ((cos φ_1  sin φ_2 Mg)/(sin (φ_1 −φ_2 )))=((mg)/2)  if M=m as given  ((cos φ_1  sin φ_2 )/(sin (φ_1 −φ_2 )))=(1/2)  2 cos φ_1 sin φ_2 =sin φ_1  cos φ_2 −cos φ_1  sin φ_2   3 cos φ_1 sin φ_2 =sin φ_1  cos φ_2   3 tan φ_2 =tan φ_1   ⇒3μ_2 =μ_1 =3 tan^(−1) (θ/2)
tanϕ1=μ1tanϕ2=μ2θ=2ϕ2ϕ2=θ2μ2=tan1ϕ2=tan1θ2ϕ1>ϕ2μ1>μ2R1sinϕ2=R2sinϕ1=Mgsin(ϕ1ϕ2)R1=sinϕ2Mgsin(ϕ1ϕ2)R2=sinϕ1Mgsin(ϕ1ϕ2)N1=R1cosϕ1=mg2cosϕ1sinϕ2Mgsin(ϕ1ϕ2)=mg2ifM=masgivencosϕ1sinϕ2sin(ϕ1ϕ2)=122cosϕ1sinϕ2=sinϕ1cosϕ2cosϕ1sinϕ23cosϕ1sinϕ2=sinϕ1cosϕ23tanϕ2=tanϕ13μ2=μ1=3tan1θ2
Commented by mr W last updated on 15/Mar/25
what did they say as you told them  that the question is wrong and  therefore their answer is wrong too?
whatdidtheysayasyoutoldthemthatthequestioniswrongandthereforetheiransweriswrongtoo?
Commented by Tawa11 last updated on 15/Mar/25
They admit the question is not well  contructed sir.  The lecturer said he will change it.  Thanks for your time sir.
Theyadmitthequestionisnotwellcontructedsir.Thelecturersaidhewillchangeit.Thanksforyourtimesir.
Commented by mr W last updated on 15/Mar/25
so i′m right. but i wonder how they  could get that answer ((3 sin θ)/(1+cos θ))  nevertheless.
soimright.butiwonderhowtheycouldgetthatanswer3sinθ1+cosθnevertheless.
Answered by mr W last updated on 14/Mar/25
if the question is modified like  following, then it can be solved.
ifthequestionismodifiedlikefollowing,thenitcanbesolved.
Commented by mr W last updated on 14/Mar/25
Commented by mr W last updated on 15/Mar/25
Commented by mr W last updated on 15/Mar/25
tan φ=μ  with 0<ϕ<(π/2)  (R_1 /(sin (θ−φ)))=(R_2 /(sin (ϕ−θ+2φ)))=((Mg)/(sin (ϕ−φ)))  R_1 =((Mg sin (θ−φ))/(sin (ϕ−φ)))  R_1 cos (ϕ+φ)=((mg)/2)  ((Mg sin (θ−φ) cos (ϕ+φ))/(sin (ϕ−φ)))=((mg)/2)  since M=m,  ⇒((sin (θ−φ) cos (ϕ+φ))/(sin (ϕ−φ)))=(1/2)  2(sin θ cos φ−cos θ sin φ)(cos ϕ cos φ−sin ϕ sin φ)=sin ϕ cos φ−cos ϕ sin φ  2(tan θ−tan φ)(1−tan ϕ tan φ)=(tan ϕ−tan φ)(√((1+tan^2  θ)(1+tan^2  φ)))  ⇒2(tan θ−μ)(1−μ tan ϕ)−(tan ϕ−μ)(√((1+tan^2  θ)(1+μ^2 )))=0
tanϕ=μwith0<φ<π2R1sin(θϕ)=R2sin(φθ+2ϕ)=Mgsin(φϕ)R1=Mgsin(θϕ)sin(φϕ)R1cos(φ+ϕ)=mg2Mgsin(θϕ)cos(φ+ϕ)sin(φϕ)=mg2sinceM=m,sin(θϕ)cos(φ+ϕ)sin(φϕ)=122(sinθcosϕcosθsinϕ)(cosφcosϕsinφsinϕ)=sinφcosϕcosφsinϕ2(tanθtanϕ)(1tanφtanϕ)=(tanφtanϕ)(1+tan2θ)(1+tan2ϕ)2(tanθμ)(1μtanφ)(tanφμ)(1+tan2θ)(1+μ2)=0
Commented by Tawa11 last updated on 14/Mar/25
Have been waiting for something to  drop here, as I saw empty space.  Hahahahahaha
Havebeenwaitingforsomethingtodrophere,asIsawemptyspace.Hahahahahaha
Commented by Tawa11 last updated on 14/Mar/25
Thanks sir.  I really appreciate.
Thankssir.Ireallyappreciate.
Commented by mr W last updated on 14/Mar/25
right or not?
rightornot?
Commented by Tawa11 last updated on 14/Mar/25
Right sir.
Rightsir.

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